The decopostion of formic acid follows first-order kinetics.
HCO2H(g)==>CO2(g) + H2(g)
The half-life for the reaction at 550 degrees Celsius is 24 seconds. How many seconds does it take for the formic acid concentration to decrease by 75%?
Calculate k = 0.693/t(1/2) = 0.693/24
Then ln(No/N) = kt
No = 100 to start (any number will do)
N = reduced by 75% so N = 25?
Plug in N, No, and k, solve for t.
Post your work if you get stuck.
Where does the 0.693 come from?
To determine the time it takes for the formic acid concentration to decrease by 75%, we can use the formula for first-order reactions:
ln(C₀/C) = kt
Where:
C₀ is the initial concentration of formic acid
C is the final concentration of formic acid
k is the rate constant
t is time
Since we are given the half-life for the reaction, we can use it to find the rate constant:
t½ = 0.693/k
Given that the half-life is 24 seconds, we can rearrange the equation to solve for k:
k = 0.693/t½
k = 0.693/24
k ≈ 0.028875 s⁻¹
Now, let's substitute the known values into the formula and solve for time:
ln(C₀/C) = kt
ln(1/0.25) = (0.028875 s⁻¹)(t)
ln(4) = (0.028875 s⁻¹)(t)
Using natural logarithm properties, we can solve for t:
t = ln(4)/(0.028875 s⁻¹)
t ≈ 4.584 s
Therefore, it takes approximately 4.584 seconds for the formic acid concentration to decrease by 75%.
To solve this problem, we need to use the concept of half-life and the first-order reaction kinetics.
The half-life of a reaction is the amount of time it takes for the concentration of the reactant to decrease by half. In this case, we are given that the half-life of the reaction is 24 seconds.
We can use the half-life formula for a first-order reaction to find the rate constant (k):
t1/2 = (0.693 / k)
where t1/2 is the half-life and k is the rate constant.
Plugging in the given half-life value:
24 seconds = (0.693 / k)
Now we can solve for the rate constant (k):
k = 0.693 / 24
k ≈ 0.0289 s^-1
Now that we have the rate constant, we can use it to determine the time it takes for the concentration of the reactant to decrease by 75%.
The general equation for the concentration of a first-order reaction over time (t) is given by:
[C] = [C0] * e^(-kt)
where [C] is the concentration at time t, [C0] is the initial concentration, k is the rate constant, and e is the mathematical constant (approximately equal to 2.71828).
We are asked to find the time it takes for the concentration to decrease by 75%. This means we need to find the time (t) when the concentration ([C]) is equal to 0.25 times the initial concentration ([C0]). Since we know k, we can plug in the values:
0.25[C0] = [C0] * e^(-kt)
Simplifying the equation:
0.25 = e^(-kt)
Taking the natural logarithm (ln) of both sides:
ln(0.25) = -kt
Now we can solve for t:
t = -ln(0.25) / k
Plugging in the value for k that we found earlier:
t ≈ -ln(0.25) / 0.0289
Using a calculator, we find:
t ≈ 96.38 seconds
Therefore, it takes approximately 96.38 seconds for the formic acid concentration to decrease by 75%.